Tutorial 7: Stokes Theorem
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- Suppose and is the part of the sphere below the plane , oriented with the outward-pointing normal (so that the normal at is ). Compute the flux integral using Stokeโs theorem.
Solution
Again we integrate the line integral over the boundary curve rather than the flux integral over the (more complicated) surface .
The boundary curve is the circle (or ) in the plane , but a note of caution is in order.
The natural parameterization (or the one we usually think of) is actually parameterizes (that is, with the opposite orientation)!
Why is that? Imagine a person walking this boundary with their head in the normal (outward) direction. The remaining part of the sphere is on their right if theyโre walking counter-clockwise. It should be on their left, so they should be walking clockwise.
Weโll calculate anyway, since we like the parameterisation. In terms of this parametrisation,
Thus,
and so, .
- Use Stokeโs theorem to evaluate where and is the part of above the . Assume that is oriented upwards.
Solution
In this case the boundary curve will be where the surface intersects the plane and so will be the curve
So, the boundary curve will be the circle of radius 2 that is in the planeย . The parameterization of this curve is,
The first two components give the circle and the third component makes sure that it is in the planeย .
Using Stokesโ Theorem we can write the surface integral as the following line integral.
So, it looks like we need a couple of quantities before we do this integral. Letโs first get the vector field evaluated on the curve. Remember that this is simply plugging the components of the parameterization into the vector field.
Next, we need the derivative of the parameterization and the dot product of this and the vector field.
We can now do the integral.
- Use Stokeโs theorem to evaluate where and is the triangle with vertices , and with counter clockwise rotation.
Solution
We are going to need the curl of the vector field eventually so letโs get that out of the way first.
Now, all we have is the boundary curve for the surface that weโll need to use in the surface integral. However, as noted above all we need is any surface that has this as its boundary curve. So, letโs use the following plane with upwards orientation for the surface.
Since the plane is oriented upwards this induces the positive direction onย ย as shown. The equation of this plane is,
Now, letโs use Stokesโ Theorem and get the surface integral set up.
Okay, we now need to find a couple of quantities. First letโs get the gradient. Recall that this comes from the function of the surface.
Note as well that this also points upwards and so we have the correct direction.
Now,ย ย is the region in theย -plane shown below,
We get the equation of the line by plugging inย into the equation of the plane. So based on this the ranges that defineย are,
The integral is then,
Donโt forget to plug in forย since we are doing the surface integral on the plane. Finishing this out gives,
- Verify Stokeโs Theorem for the field on the ellipse .
Solution
We compute both sides in
We start computing the circulation integral on the ellipse . We need to choose a counterclockwise parametrization, hence the normal to points upwards.
We choose, for
Therefore, the right-hand rule normal to is
The circulation integral is:
The substitution on the first term and , implies .
Since , we conclude that
We now compute the right-hand side in Stokesโ Theorem.
is the flat surface , so
Then,
The right-hand side above is twice the area of the ellipse, Since that an ellipse has area , we obtain
- Verify Stokeโs Theorem for is ๐ is the paraboloid with the circle as its boundary.
Solution
Surface integral,
is region in side , a unit circle.
Switching to polar coordinates:
Hence,
Line integral,
: unit circle โ Switch to polar coordinates,
Hence,
- Use Stokeโs Theorem to calculate for where is the part of the ellipsoid below the -plane and is the lower normal.
Solution
Using polar coordinates,
Hence,
- Use Stokeโs Theorem to evaluate the line integral where is the vector field and is the curve of intersection of the cylinder and the plane and is oriented in a counterclockwise direction when viewed from above.
Solution
The of is computed as,
Now, writing the plane as the level surface
Applying Stokesโ theorem,
- Evaluate the line integral of over the curve that is the intersection of the cylinder with the plane .
Solution
To describe the surface enclosed by , we use the parameterisation
Using and , we obtain,
Compute the curl,
Let be the domain of the parameter,
Using Greenโs Theorem,
- Evaluate where and is the part of the sphere that lies inside the cylinder and above the -plane.
Solution
Surface is bounded by a circle formed by the intersection of the sphere of radus 2 and the cylinder of radius 1.
We can describe using the vector-valued function,
Using Stokesโ Theorem,